why does parfor execute loops in a random order?
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I can't see any reason why, and if my PC crashes during a 20,000 loop iteration, it's a pain to figure out which iterations are still to be done.
Cheers Mike
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Jason Ross
el 28 de Feb. de 2013
Just to be clear -- parfor loops are of independent of iteration order, and do not guarantee deterministic results. It's not random.
The functional reason for this is if iteration 2 finishes before iteration 1 then iteration 3 can start work, and so on, since each iteration is independent of the other.
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Jason Ross
el 1 de Mzo. de 2013
You might also want to investigate using the job/task interface, which gives you more control over the iterations and tracking down errors. A simple example:
c=parcluster();
alloutputs = [];
for ii=1:10
job = c.createJob;
for jj=1:10
createTask(job,@rand, 1, {3,3});
end
job.submit;
job.wait;
outputs = job.fetchOutputs;
alloutputs = [alloutputs ; outputs];
disp(ii);
job.destroy;
end
Note that things like preallocation, fancy formatting and error checking have been left out. Things that will likely be interesting to you:
- The job object contains the Task ID of errors. You could use this information to know what failed.
- You will likely need to figure out what work for you in terms of iteration display, number of tasks per job, dealing with the returns, etc.
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